Question

\(\lim\limits_{x\rightarrow1}\)-\(\frac{x^3-1}{|1-x|}\)

Answer 1

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1}\frac{x^3-1}{x-1}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}=\lim\limits_{x\rightarrow1}\left(x^2+x+1\right)=3\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\frac{x^3-1}{1-x}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2+x+1\right)}{-\left(x-1\right)}=-3\)

\(\Rightarrow\lim\limits_{x\rightarrow1^+}f\left(x\right)\ne\lim\limits_{x\rightarrow1^-}f\left(x\right)\)

\(\Rightarrow\) Không tồn tại \(\lim\limits_{x\rightarrow1}\frac{x^3-1}{\left|1-x\right|}\)